Question
SHow Work
Cloud drop growth by vapor deposition. The growth rate of a cloud drop can be approximated by the expression: rd (t) = [2G(s - sk)t + rdo2]1/2 where G = Dv Mw es (T infinity) / rho1 R* Tinfinity [1 + Dv nsinfinity lv2 / kT R* Tinfinity2] where rd is the drop radius.mum), Dv is the diffusion coefficient of water vapor molecules through air (Dv = 2.11 Times 10-5 (T/273)2(1013/p) m2 s-1), Mw is the water molar mass, rho1 is the liquid water density, lv is the molar specific enthalpy of vaporization, kT is the thermal conductivity of dry air (kT ~ 2.2 Times 10-2 J m-1 s-1 K-1), nsinfinity is the molar concentration of saturated water vapor at temperature Tinfinity. G is usually given the units of mum2 s-1 Verify that the units for G and the units for rd(t) are correct. You will need to change all the mass-based constants, like lv, that you are used to using to molar based - amount or number per mole, not kg. For Tinfinity = 283 K, what are es(Tinfinity) and nsinfinity? What is the numerical value for G? For a drop that starts with an aerosol of radius 0.05 mum, plot the radius increase as a function of time. Assume that s = 0.005 and sk ~ 0.001. Let time be between 0 and 2000 seconds. How long will it take for a drop to grow by vapor deposition to a size of 1 mm?
Explanation / Answer
a) The units can be verified..its simple enough. b) When T = 283K e(t=infinity) = 45x10-3 and n(s=infinity) = 0.582 c) G = 23,76x10-2 um^2s^-1. Good luck.
